Problem: Is ${68549}$ divisible by $3$ ?
Explanation: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {68549}= &&{6}\cdot10000+ \\&&{8}\cdot1000+ \\&&{5}\cdot100+ \\&&{4}\cdot10+ \\&&{9}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {68549}= &&{6}(9999+1)+ \\&&{8}(999+1)+ \\&&{5}(99+1)+ \\&&{4}(9+1)+ \\&&{9} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {68549}= &&\gray{6\cdot9999}+ \\&&\gray{8\cdot999}+ \\&&\gray{5\cdot99}+ \\&&\gray{4\cdot9}+ \\&& {6}+{8}+{5}+{4}+{9} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first four terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${68549}$ is divisible by $3$ if ${ 6}+{8}+{5}+{4}+{9}$ is divisible by $3$ Add the digits of ${68549}$ $ {6}+{8}+{5}+{4}+{9} = {32} $ If ${32}$ is divisible by $3$ , then ${68549}$ must also be divisible by $3$ ${32}$ is not divisible by $3$, therefore ${68549}$ must not be divisible by $3$.